is $R(T_1+T_2)=R(T_1)+R(T_2)$ true?

Question

If $T_1,T_2\in\mathcal{L}(H)$ be two linear bounded operators on the Hilbert space $H= H_1\oplus H_2$ with $R(T_1)\subseteq H_1 $ and $R(T_2)\subseteq H_2 $ ,

is $R(T_1+T_2)=R(T_1)+R(T_2)$ true ?, where R(T) represents the range of T. If no, then under what condition, my argument will be true.

Here $R(T_1+T_2)\subseteq R(T_1)+R(T_2).$ I am unable to prove or disprove the converse. Kindly provide suggestions.

Thanks.

Answer 1

Let $H=\mathbb C\oplus\mathbb C$, and $$ T_1=\begin{bmatrix} 1&1\\0&0\end{bmatrix},\qquad T_2=\begin{bmatrix}0&0\\1&1\end{bmatrix}. $$ Then $R(T_1)=\mathbb C_1\oplus 0$ and $R(T_2)=0\oplus\mathbb C$. So $R(T_1)+R(T_2)=H$, while $T_1+T_2$ is rank-one. Thus $R(T_1+T_2)\subsetneq R(T_1)+R(T_2)$.

Answer 2

As @Brian Moehring suggested, it's worthwhile analyzing the containment $$R(T_1)+R(T_2)\subset R(T_1+T_2)\quad (1)$$ It holds iff $$H=\ker T_1+\ker T_2 \qquad (2)$$ Indeed if (2) holds then $$R(T_1)=T_1(H)=T_1(\ker T_2)=(T_1+T_2)(\ker T_2)\subset R(T_1+T_2)$$ Similarly $R(T_2)\subset R(T_2+T_2).$

Assume (1) holds. Then for any $x\in H$ we have $$T_1x=(T_1+T_2)y$$ for some y. Hence $$T_1(x-y)=T_2y$$ and both terms must vanish. Thus $y\in \ker T_2$ and $x-y\in \ker T_1.$ Therefore $$x=(x-y)+y\in \ker T_1+\ker T_2$$

In particular (1) does not hold if $T_1,T_2\neq 0$ and $\ker T_1\subset\ker T_2.$ Such example can be constructed as follows. Let $T:H_0\to H_0$ be an invertible operator. Consider $H=H_0\oplus H_0$ and the operators $$T_1=\begin{pmatrix} T &-T\\ 0 &0\end{pmatrix},\quad T_2=\begin{pmatrix} 0 &0\\ T&-T\end{pmatrix}$$ Then $$\ker T_1=\ker T_2=\{(x,x)\,:\, x\in H_0\}$$